Independent solution

How to solve this Exponential Distribution question

Answer in brief

For an exponential lifetime with mean θ, the variance is θ² and the percentile rank of a value x is 100F(x). Evaluating the exponential CDF at x = θ² gives 100(1 - e^{-θ}), which is choice E.

Setup

Setup

Write the distribution function and variance using the mean parameter.

F(x)=1−e−x/θ(x≥0),Var⁡(X)=θ2F(x)=1-e^{-x/\theta}\quad(x\ge 0),\qquad \operatorname{Var}(X)=\theta^2

Model

Model

The requested percentile rank is the distribution function evaluated at the stated lifetime value, multiplied by 100.

k100=F(θ2)\frac{k}{100}=F(\theta^2)

Compute

Compute

Substitute θ² into the exponential CDF and simplify its exponent.

F(θ2)=1−e−θ2/θ=1−e−θF(\theta^2)=1-e^{-\theta^2/\theta}=1-e^{-\theta}
k=100(1−e−θ)k=100\left(1-e^{-\theta}\right)

Answer

Answer

The symbolic expression matches choice E.

k=100(1−e−θ)(E)\boxed{k=100\left(1-e^{-\theta}\right)\quad\text{(E)}}